Again, turn to Example 4, on page 57, and let us investigate the reciprocal theorem. The three conics touching respectively each pair of the sides of a triangle at the angular points where they meet the third side, will reciprocate into "three conics passing respectively through each pair of the angular points of a triangle, and touching the lines joining them with the third angular point," that is, the sides of the triangle themselves. This condition, therefore, reciprocates into itself. The condition "all intersecting in a point" reciprocates into "all touching a straight line." Hence the data are, "Three conics are drawn, touching respectively each pair of the angular points of the sides of a triangle at the points where they meet the third side, and all touching a straight line."

In the matter to be proved, we may first enquire what are the reciprocals of "the sides of the triangle which intersect" (that is, which do not touch) "their respective conics." These will be "the angular points of the triangle not lying on their respective conics." The three tangents at their common point will reciprocate into "the three points of contact of their common tangents." And the meeting of the tangents with the sides will reciprocate into the lines joining the points of contact with the angular points. Hence the first thing to be proved is, "That the three straight lines joining the points of contact of the common tangent with the angu lar points of the triangle not lying on the respective conics all pass through a point.'

Again, "the other common tangents to each pair of conics" reciprocate into "the other points of intersection of each pair of conics," and "the sides of the triangle which touch the several pairs of conics" into the angular points of the triangle "common to the several pairs of conics." Hence the latter part of the theorem will run: And that the same three straight lines respectively join the other point of intersection of each pair of conics with the angular point of the triangle common to each pair."

66

straight line is denoted by a single letter, and a point by the pair of letters representing any two straight lines which intersect in it, a useful mode of familiarizing himself with the method of reciprocal polars.

TRANSFORMATION OF THEOREMS.

111

11. After a little practice, the process of reciprocating a given theorem will be found to consist simply in writing "straight line" for "point," "join" for "intersect," "locus for " envelope," &c., and vice versâ. The word "conic" will of course remain unaltered.

12. Brianchon's Theorem.

By reciprocating Pascal's Theorem (given in Art. 12, Chap. III.), we obtain Brianchon's Theorem, which asserts that

"If a hexagon be described about a conic section, the three diagonals will intersect in a point*.

* It may be well to append an independent proof of this important theorem. Take three sides of the hexagon as lines of reference, and let the equations of the other three be

a+m ̧ß+n ̧y=0, l2a+ẞ+n2y=0, lзa+m2ß+y=0.

Let the equation of the conic be

The line passing through the intersections of B=0 with (l3, m3, 1) and of y=0 with (l,, 1, n) is represented by the equation

+ y = 0;

The student will find it useful to transform, by the method of reciprocal polars, the special cases of Pascal's Theorem, given in Art. 13, Chap. III.; and to obtain a geometrical construction by which when five tangents to a conic are given, their points of contact may be found.

13. The anharmonic ratio of the pencil formed by four intersecting straight lines is the same as that of the range formed by their poles. This may be proved as follows.

Let OP, OQ, OR, OS be the four straight lines, P', Q, R', S' their poles, which will lie in a straight line, the polar

OA

Q'R' S'

Fig. 18.

of O; let P, Q, R, S be the points in which the pencil is cut by the transversal P'Q'R'S'.

Let this transversal cut the conic in K, K. Bisect

ANHARMONIC PROPERTIES.

113

KK in V. Then, since PKPK, is divided harmonically in ̃P‚“K ̧‚ P'‚ K, (Art. 21, Chap. 11.), it follows that

whence

2

PK1. P'K ̧=PK.P'K,

(VP− VK,) (VK,+VP)=(VP+VK,) (VK, – VP),

which since VK1 = VK,, reduces to

VP.VP=VK,

which, by similar reasoning,

=

VQ. VQ' VR. VR' = VS. VS".

=

Hence the eight points P, Q, R, S, P, Q, R, S' are in an involution, of which K1, K2 are the foci, and therefore (Art. 27, Chap. I.)

2

{0.PQRS} = [P' Q'R' S′].

14. In Art. 13, Chap. I. we saw that the condition that the three points (l1, m1, n1), (l,, m,, n), (1 ̧, m, n,) shall lie in the same straight line is identical with the condition that the three straight lines (l, m,, n1) (l2, m2, n2), (13, m3, n3) shall intersect in the same point. Now these several points and lines are respectively the poles and polars of each other, with respect to the imaginary conic

a2 + B2 + y2 = 0.

Thus the theory of reciprocal polars explains the fact that the condition for three points lying in a straight line is identical with that for three straight lines intersecting in a point. It also explains the identity of conditions noticed in Chap. II. Arts. 7 and 9.

For the reciprocal of the conic

λ2a2 +μ3ß2 + v3y2 — 2μvßy — 2vλya - 2λμaß = 0...... (1),

with respect to,

will be found to be

C

And the polar of (f, g, h) is fa+gB + hy = 0.

Hence if the line fa+gB+hy = 0 touch (1), the point (f, g, h) lies in (2), giving for the condition of tangency

And if the line fa+gB+hy = 0 touch (2), the point (f, g, h) lies in (2), giving for the condition of tangency in

that case

λ2ƒa2 + μ3g2 + v3h2 — 2μvgh — 2vλhƒ — 2λμfg = 0.

These conditions of tangency are identical with those already investigated.

Again, every parabola touches the line at infinity. Now the co-ordinates of the pole of this line are proportional to a, b, c. Hence, if the conic, represented by the general equation of the second degree, be a parabola, the point (a, b, c) must lie in the reciprocal conic. This gives, as the condition for a parabola,

Ua2 + Vb2 + Wc2 + 2 U'bc + 2 V'ca + 2 W'ab = 0,

the same as that already investigated.

15. PROP. Any straight line drawn through a given point A is divided harmonically by any conic section, and the polar of A with respect to it.

This proposition may be proved as follows. Let the straight line cut the curve in P and Q, and the polar of A in B. Let C be the polar of the straight line, and let ABC be the triangle of reference. The conic will be self-conjugate with respect to ABC, and will therefore be represented by the equation

ua2 + vß2 + wy2 = 0.

Hence the lines CP, CQ, which are tangents to the conic, are represented by the equation

ua2 + vẞ2 = 0,

and therefore form an harmonic pencil with CA, CB.